# consider the group $G=\mathbb Q/\mathbb Z$. For $n>0$, is there a cyclic subgroup of order n

consider the group $G=\mathbb Q/\mathbb Z$. Let $n$ be a positive integer. Then is there a cyclic subgroup of order $n$?

1. not necessarily.
2. yes, a unique one.
3. yes, but not necessarily a unique one.
4. never
-
i have the idea that Q is not cyclic but how can i use this? – poton Aug 31 '12 at 16:43
Perplexed? Here is a guide to help you: think of $\mathbb{Q}/\mathbb{Z}$ as the group of all roots of unity in $\mathbb{C}$. Why are these two groups isomorphic? Map $\mathbb{Q}$ into the complex numbers by $\lambda\mapsto e^{2\pi i\lambda}$, clearly a homomorphism from $\mathbb{Q}$ as additive group to the multiplicative group of complexes, and with image equal to the group of roots of unity. What’s the kernel? – Lubin Aug 31 '12 at 17:17

Another hint to prove unicity: suppose $\overline {a/b}$ has order $n$, $0 < a < b$, and $gcd(a,b)=1$, then $na=kb$ for some $k \in \mathbb{Z}$, hence $a$ divides $k$ and $n=bk'$. It follows that $\overline {a/b}$ = $\overline {ak'/n}$, so ...
Hint: $\dfrac{1}{n}$ ${}{}{}{}{}{}{}{}{}$
Yes, the group is unique. For me, because of my number-theoretic inclinations, the reason is that if $m \gt 0$, and $a$ and $m$ are relatively prime, then the numbers $a$, $2a$, $3a$, and so on up to $(m-1)a$ are incongruent modulo $m$, so one of them is congruent to $1$ modulo $m$, say $qa$. Then $qa/\mathbb{Z}=1/\mathbb{Z}$. – André Nicolas Aug 31 '12 at 17:12